The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is 20 cm, the length of the open organ pipe is

Options

A. `13.2 cm`
B. `8 cm`
C. `12.5 cm`
D. `16 cm`

Explanation

To solve the problem, we need to find the length of the open organ pipe (L₀) given that the fundamental frequency of the open organ pipe is equal to the third harmonic of a closed organ pipe, and the length of the closed organ pipe (Lₐ) is 20 cm.

### Step-by-Step Solution:

1. **Understand the Frequencies**:
- The fundamental frequency (f₀) of an open organ pipe is given by the formula:
\[ f₀ = \frac{V}{2L₀} \]
- The frequency of the nth harmonic of a closed organ pipe is given by:
\[ f_n = \frac{nV}{4Lₐ} \]
- For the third harmonic (n = 3), the formula becomes:
\[ f_3 = \frac{3V}{4Lₐ} \]

2. **Set the Frequencies Equal**:
- According to the problem, the fundamental frequency of the open pipe is equal to the third harmonic of the closed pipe:
\[ f₀ = f_3 \]
- Therefore, we can write:
\[ \frac{V}{2L₀} = \frac{3V}{4Lₐ} \]

3. **Cancel the Speed of Sound (V)**:
- Since V appears in both sides of the equation, we can cancel it out:
\[ \frac{1}{2L₀} = \frac{3}{4Lₐ} \]

4. **Cross Multiply**:
- Cross-multiplying gives us:
\[ 4Lₐ = 6L₀ \]

5. **Rearranging for L₀**:
- Rearranging the equation to solve for L₀:
\[ L₀ = \frac{4Lₐ}{6} = \frac{2Lₐ}{3} \]

6. **Substituting the Length of the Closed Pipe**:
- We know that the length of the closed organ pipe (Lₐ) is 20 cm:
\[ L₀ = \frac{2 \times 20 \text{ cm}}{3} = \frac{40 \text{ cm}}{3} \]

7. **Calculating the Final Length**:
- Performing the division:
\[ L₀ = 13.33 \text{ cm} \]

### Final Answer:
The length of the open organ pipe is approximately \( \frac{40}{3} \) cm, which is about 13.33 cm.